Elements of Chapter 9: Nonlinear Systems To solve x0 = Ax, we use the ansatz that x(t) = eλtv. We found that λ is an eigenvalue of A, and v an associated eigenvector. We can also summarize the geometric behavior of the solutions by looking at a plot- However, there is an easier way to classify the stability of the origin (as an equilibrium), To find the eigenvalues, we compute the characteristic equation: p Tr(A) ± ∆ λ2 − Tr(A)λ + det(A) = 0 λ = 2 which depends on the discriminant ∆: • ∆ > 0: Real λ1; λ2. • ∆ < 0: Complex λ = a + ib • ∆ = 0: One eigenvalue. The type of solution depends on ∆, and in particular, where ∆ = 0: ∆ = 0 ) 0 = (Tr(A))2 − 4det(A) This is a parabola in the (Tr(A); det(A)) coordinate system, inside the parabola is where ∆ < 0 (complex roots), and outside the parabola is where ∆ > 0. We can then locate the position of our particular trace and determinant using the Poincar´eDiagram and it will tell us what the stability will be. Examples Given the system where x0 = Ax for each matrix A below, classify the origin using the Poincar´eDiagram: 1 −4 1. 4 −7 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = −6 Det(A) = −7 + 16 = 9 ∆ = 36 − 4 · 9 = 0 Therefore, we have a \degenerate sink" at the origin. 1 2 2. −5 −1 SOLUTION: Compute the trace, determinant and discriminant: Tr(A) = 0 Det(A) = −1 + 10 = 9 ∆ = 02 − 4 · 9 = −36 The origin is a center. 1 3. Given the system x0 = Ax where the matrix A depends on α, describe how the equilibrium solution changes depending on α (use the Poincar´e Diagram): 2 −5 (a) α −2 SOLUTION: The trace is 0, so that puts us on the \det(A)" axis. The determinant is −4 + 5α. If this is positive, we have a center: 4 −4 + 5α > 0 ) α > ) The origin is a CENTER 5 4 α < ) The origin is a SADDLE 5 4 If α = 5 , we have \uniform motion". That is, x1(t) and x2(t) will be linear in t (see if you can find the general solution!). α 1 (b) −1 α SOLUTION: The trace is 2α and the discriminant is α2 + 1. The discriminant is: ∆ = 4α2 − 4(α2 + 1) = 4α2 − 4α2 − 4 = −4 Therefore, we are always inside the parabola in the upper part of the graph, so the sign of the trace will tell us if we have a SPIRAL SINK (α < 0), a CENTER (α = 0), or a SPIRAL SOURCE (α > 0). 4. In addition to classifying the origin, find the general solution to the system x0 = Ax using eigenvalues and eigenvectors for the matrix A below. −1 −4 A = 1 −1 SOLUTION: The trace is −2 and the determinant is 5, and the dis- criminant is 4 − 4 · 5 = −16, so the origin is a SPIRAL SINK. The characteristic equation is λ2 + 2λ + 5 = 0 ) λ2 + 2λ + 1 + 4 = 0 ) (λ + 1)2 = −4 and λ = −1 ± 2i. For λ = −1 + 2i, we find the corresponding eigen- vector: (−1 + 1 − 2i)v − 4v = 0 v = 2iv 2i 1 2 ) 1 2 ) v = v1 + (−1 + 1 − 2i)v2 = 0 v2 = v2 1 Now we compute eλtv: 2i 2i −2 sin(2t) + i2 cos(2t) e(−1+2i)t = e−t(cos(2t)+i sin(2t)) = e−t 1 1 cos(2t) + i sin(t) 2 Now the solution to the differential equation is: λt λt x(t) = C1Re(e v) + C2Im(e v) The exponential can be factored out to make it simpler to write: −2 sin(t) 2 cos(t) x(t) = e−t C + C 1 cos(2t) 2 sin(2t) Just for fun, we could solve this last system using Maple. Just type in: sys_ode := diff(x(t),t)= -x(t)-4*y(t), diff(y(t),t)= x(t)-y(t); dsolve([sys_ode],[x,y]); Linearizing a Nonlinear System The following notes are elements from Sections 9.2 and 9.3. • Suppose we have an autonomous system of equations: x0 = f(x; y) y0 = g(x; y) Then (as before) we define a point (a; b) to be an equilibrium point for the system if f(a; b) = 0 AND g(a; b) = 0 (that is, you must solve the system of equations, not one at a time). • Example: Find the equilibria to: x0 = −(x − y)(1 − x − y) y0 = x(2 + y) SOLUTION: From the second equation, either x = 0 or y = −2. Take each case separately. { If x = 0, then the first equation becomes y(1 − y), so y = 0 or y = 1. So far, we have two equilibria: (0; 0) (0; 1) { Next, if y = −2 in the second equation, then the first equation becomes −(x + 2)(1 − x + 2) = 0 ) x = −2 or x = 3 We now have two more equilibria: (−2; −2) (3; −2) 3 • Key Idea: The \interesting" behavior of a dynamical system is orga- nized around its equilibrium solutions. • To see what this means, here is the graph of the direction field for the example nonlinear system: • In order to understand this picture, we will need to linearize the differ- ential equation about its equilibrium. • Let x = a; y = b be an equilibrium solution to x0 = f(x; y) and y0 = g(x; y). Then the linearization about (a; b) is the system: 0 u fx(a; b) fy(a; b) u 0 = v gx(a; b) gy(a; b) v where u = x − a and y = v − b. In our analysis, we really only care about this matrix- You may have used it before, it is called the Jacobian matrix. • Continuing with our previous example, we compute the Jacobian ma- trix, then we will insert the equilibria one at a time and perform our local analysis. We then try to put together a global picture of what's happening. Recall that the system is: x0 = −(x − y)(1 − x − y) = −x + x2 + y − y2 y0 = x(2 + y) = 2x + xy The Jacobian matrix for our example is: f f −1 + 2x 1 − 2y x y = gx gy 2 + y x 4 Equilibrium System Tr(A) det(A) ∆ Poincare −1 1 (0; 0) −1 −2 Saddle 2 0 −1 −1 Spiral (0; 1) −1 3 −11 3 0 Sink −5 5 (−2; −2) −7 10 9 Sink 0 −2 5 5 (3; −2) 8 15 4 Source 0 3 Here's the picture again: 5 System Tr(A) det(A) ∆ Poincare λ V Fill in 1 2 2 0 9 −36 3i −5 −1 −1 + 3i 1 −1 −1 0 4 4 0 2; 2 1 3 1 1 1 − 2 1 1 1 1 −1 5=4 −4 − 2 + i −1 − 2 i −1 −1 1 −4 1 −5=4 1=4 9=16 −1; −1=4 0 − 4 0 3 Figure 1: From top to bottom, Center, Degenerate Source, Spiral Sink, Sink. This is for class discussion. 6 Homework: Elements of Chapter 9, Day 1 1. Fill in the following and under \Poincar´e"classify the origin. Then, given the eigenvalues/eigenvectors, also write down the general solution to x0 = Ax. In the case that there is only one eigenvector, the second column of V shows the generalized eigenvector w. System Tr(A) det(A) ∆ Poincare λ V Fill in 3 −2 1 1 + 2i 4 −1 1 − i 2 −1 1 1 −1; 1 3 −2 3 1 0 2 −i 2i −2 0 1 4 −2 1 0 0; 0 8 −4 2 −1=2 2. Explain how the classification of the origin changes by changing the α in the system: 0 α (a) x0 = x 1 −2 2 α (b) x0 = x 1 −1 α 10 (c) x0 = x −1 −4 Hint: Use a number line to keep track of where the trace, deter- minant and discriminant change sign. 3. For the following nonlinear systems, find the equilibrium solutions (the derivatives are with respect to t, as usual). (a) x0 = x − xy, y0 = y + 2xy (b) x0 = y(2 − x − y), y0 = −x − y − 2xy (c) x0 = 1 + 2y, y0 = 1 − 3x2 4. For each of the systems in the previous problem, find the Jacobian ma- trix, then linearize about each equilibrium. Use the Poincar´eDiagram to classify each equilibrium solution. 7.